DC Circuits - How to Determine a Bulb's Brightness

[jkface]

Homework Statement

The circuit has identical bulbs and an ideal battery. When bulb D is unscrewed from its socket (the socket remains in its place), what would happen to the brightness of bulb F?

Homework Equations

V = IR

The Attempt at a Solution

Since bulb D is unscrewed from it socket and the socket remains in its place, we can replace bulb D with a switch that is open. Assigning resistance of 1 for each bulb, the right side of circuit has 2R and the current for bulb F would be I = V/2R. When the bulb D is back in its place, the resistance of the right side of the circuit would be 5R/3 and the current for bulb F would be I = (3V/5R)/(2/3) = 9V/10R. This shows that when bulb D is unscrewed from its socket, bulb F would have less brightness.

When I entered the above explanation it turned out to be wrong. Can anyone explain why the brightness of the bulb increases?

 

[ehild]

When the bulb D is back in its place, the resistance of the right side of the circuit would be 5R/3 and the current for bulb F would be I = (3V/5R)/(2/3) = 9V/10R.

Check the current through F. Can it be higher than the main current?

ehild

 

[PeterO]

Homework Statement

The circuit has identical bulbs and an ideal battery. When bulb D is unscrewed from its socket (the socket remains in its place), what would happen to the brightness of bulb F?

Homework Equations

V = IR

The Attempt at a Solution

Since bulb D is unscrewed from it socket and the socket remains in its place, we can replace bulb D with a switch that is open. Assigning resistance of 1 for each bulb, the right side of circuit has 2R and the current for bulb F would be I = V/2R. When the bulb D is back in its place, the resistance of the right side of the circuit would be 5R/3 and the current for bulb F would be I = (3V/5R)/(2/3) = 9V/10R. This shows that when bulb D is unscrewed from its socket, bulb F would have less brightness.

When I entered the above explanation it turned out to be wrong. Can anyone explain why the brightness of the bulb increases?

When globe D is removed, current passes through just B and F on the RHS. As they are in series, the current through each is the same - and the PD across each is thus equal, being half the Emf of the battery. [remember they are equal resistance globes]

When Globe D is/was in position, the current through B is shared through the two parallel arms, so more current flows through B compared to that through F - so the PD across B is greater than that over F, so the PD across B must be more than one half of the Emf of the cell, while the PD across F must be less than a half; so F must be duller when D is there - or if you like brighter when D is removed.

 

[jkface]

Check the current through F. Can it be higher than the main current?

ehild

Now that you mention it, I'm beginning to realize how stupid my answer sounds. If I multiply 3V/5R by 2/3, would that be the right answer?

 

[jkface]

When globe D is removed, current passes through just B and F on the RHS. As they are in series, the current through each is the same - and the PD across each is thus equal, being half the Emf of the battery. [remember they are equal resistance globes]

When Globe D is/was in position, the current through B is shared through the two parallel arms, so more current flows through B compared to that through F - so the PD across B is greater than that over F, so the PD across B must be more than one half of the Emf of the cell, while the PD across F must be less than a half; so F must be duller when D is there - or if you like brighter when D is removed.

Thank you. I think I'm beginning to understand the problem now. Would I reach the same conclusion if I only discuss voltage differences?

 

[PeterO]

Thank you. I think I'm beginning to understand the problem now. Would I reach the same conclusion if I only discuss voltage differences?

The brightness of the globe depends on the power being transformed/dissipated.

Power can be calculated from P = VI or I²R or V²/R

That 3rd form would suggest that Voltage difference would be sufficient.

 

[ehild]

Now that you mention it, I'm beginning to realize how stupid my answer sounds. If I multiply 3V/5R by 2/3, would that be the right answer?

Yes, of course.

ehild

 

[jkface]

Thanks everyone for helping me!